đề là rút gọn các biểu thức sau

nhờ mọi người giải giúp mình. cảm ơn mn nhìu

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\sqrt{5}+2}\)

\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)

b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)

=2căn 5-2-2căn 5

=-2

d: \(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)

\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}\)

\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}=\sqrt{2}\)

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a: \(\dfrac{6}{2-\sqrt{10}}-\dfrac{2\sqrt{5}-5\sqrt{2}}{\sqrt{2}-\sqrt{5}}+\sqrt{49+4\sqrt{10}}\)

\(=\dfrac{6\left(2+\sqrt{10}\right)}{4-10}-\dfrac{\sqrt{10}\left(\sqrt{2}-\sqrt{5}\right)}{\sqrt{2}-\sqrt{5}}+\sqrt{49+2\cdot2\sqrt{10}}\)

\(=\dfrac{6\left(2+\sqrt{10}\right)}{-6}-\sqrt{10}+\sqrt{49+2\cdot\sqrt{40}}\)

\(=-2-\sqrt{10}-\sqrt{10}+\sqrt{49+4\sqrt{10}}\)

\(=-2-2\sqrt{10}+\sqrt{49+4\sqrt{10}}\)

b: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x< >1\end{matrix}\right.\)

\(\left(\dfrac{x-\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x}\)

\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)

\(=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\cdot\dfrac{x}{\sqrt{x}+1}\)

\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{x}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\cdot\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)}\)

\(=\sqrt{x}\left(\sqrt{x}-1\right)\)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

Bài 2:

a: \(\frac{21+8\sqrt5}{4+\sqrt5}\cdot\sqrt{9-4\sqrt5}\)

\(=\frac{\left(4+\sqrt5\right)^2}{4+\sqrt5}\cdot\sqrt{\left(\sqrt5-2\right)^2}\)

\(=\left(4+\sqrt5\right)\left(\sqrt5-2\right)=4\sqrt5-8+5-2\sqrt5=2\sqrt5-3\)

b: \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

\(=\sqrt{\left(\sqrt5-\sqrt3\right)^2}-\sqrt{\left(\sqrt5+\sqrt3\right)^2}\)

\(=\sqrt5-\sqrt3-\sqrt5-\sqrt3=-2\sqrt3\)

Bài 1:

a: \(\frac{\sqrt6-\sqrt{15}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt3\left(\sqrt2-\sqrt5\right)}{-\sqrt7\left(\sqrt2-\sqrt5\right)}=-\sqrt{\frac37}=-\frac{\sqrt{21}}{7}\)

b: \(\frac{10+2\sqrt{10}}{\sqrt5+\sqrt2}+\frac{8}{1-\sqrt5}\)

\(=\frac{2\sqrt5\left(\sqrt5+\sqrt2\right)}{\sqrt5+\sqrt2}-\frac{8\left(\sqrt5+1\right)}{\left(\sqrt5-1\right)\left(\sqrt5+1\right)}\)

\(=2\sqrt5-2\left(\sqrt5+1\right)=-2\)

c: \(\frac{\sqrt{3-\sqrt5}\left(3+\sqrt5\right)}{\sqrt{10}+\sqrt2}=\frac{\sqrt{6-2\sqrt5}\left(3+\sqrt5\right)}{\sqrt{20}+2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}\cdot\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}=\frac{\left(\sqrt5-1\right)\left(3+\sqrt5\right)}{2\left(\sqrt5+1\right)}\)

\(=\frac{3\sqrt5+5-3-\sqrt5}{2\left(\sqrt5+1\right)}=\frac{2\sqrt5+2}{2\sqrt5+2}\)

=1

d: \(\frac{1}{\sqrt2+\sqrt{2+\sqrt3}}+\frac{1}{\sqrt2-\sqrt{2+\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{4+2\sqrt3}}+\frac{\sqrt2}{2-\sqrt{4+2\sqrt3}}\)

\(=\frac{\sqrt2}{2+\sqrt{\left(\sqrt3+1\right)^2}}+\frac{\sqrt2}{2-\sqrt{\left(\sqrt3-1\right)^2}}=\frac{\sqrt2}{3+\sqrt3}+\frac{\sqrt2}{3-\sqrt3}\)

\(=\frac{\sqrt2\left(3-\sqrt3\right)+\sqrt2\left(3+\sqrt3\right)}{\left(3-\sqrt3\right)\left(3+\sqrt3\right)}=\frac{3\sqrt2-\sqrt6+3\sqrt2+\sqrt6}{9-3}=\frac{6\sqrt2}{6}=\sqrt2\)

1: \(=8+2\sqrt{10}-3\sqrt{10}+\sqrt{10}=8\)

Giải giúp e chi tiết hơn được không ạ

\(\left(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{30}-\sqrt{6}}{\sqrt{5}-1}\right):\dfrac{1}{2\sqrt{5}-\sqrt{6}}\)

\(=\left(\dfrac{2\sqrt{5}\cdot\sqrt{5}+\sqrt{2}\cdot2\sqrt{5}}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{5}\cdot\sqrt{6}-\sqrt{6}\cdot1}{\sqrt{5}-1}\right):\dfrac{1}{2\sqrt{5}-\sqrt{6}}\)

\(=\left[\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{6}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right]\cdot\left(2\sqrt{5}-\sqrt{6}\right)\)

\(=\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)\)

\(=\left(2\sqrt{5}\right)^2-\left(\sqrt{6}\right)^2\)

\(=20-6\)

\(=14\)

\(=\left(\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{\sqrt{6}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\cdot\left(2\sqrt{5}-\sqrt{6}\right)\)

\(=\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)\)

=20-6

=14

bạn nên tự nghiên cứu rồi giải đi chứ bạn đưa 1 loạt thế thì ai rảnh mà giải, với lại cứ bài gì không biết chưa chịu suy nghĩ đã hỏi rồi thì tiến bộ sao được, đúng không